Last updated: 2026-07-13

What this calculator does

This calculator solves a system of three linear equations in three unknowns, x, y, and z, using Gaussian elimination:

$$a_1 x + b_1 y + c_1 z = d_1 \\ a_2 x + b_2 y + c_2 z = d_2 \\ a_3 x + b_3 y + c_3 z = d_3$$

It writes the twelve coefficients and constants as an augmented matrix, reduces that matrix to row-echelon form with row operations, and then back-substitutes to return the solution (x, y, z). When the coefficient matrix is singular it reports that the system has no unique solution.

The augmented matrix

Stacking the coefficients and the right-hand sides gives the augmented matrix, with a vertical bar separating the coefficient block from the constants:

$$\left[\begin{array}{ccc|c} a_1 & b_1 & c_1 & d_1 \\ a_2 & b_2 & c_2 & d_2 \\ a_3 & b_3 & c_3 & d_3 \end{array}\right]$$

Every step of Gaussian elimination is a row operation on this matrix: swapping two rows, scaling a row, or adding a multiple of one row to another. None of these changes the solution set.

Forward elimination

The goal of forward elimination is to create zeros below the diagonal, one column at a time, until the matrix is upper-triangular (row-echelon form).

Step 1 — clear the x-column. Using the first pivot a₁, subtract a multiple of row 1 from rows 2 and 3:

$$m_{21} = \frac{a_2}{a_1}, \quad R_2 \to R_2 - m_{21} R_1, \qquad m_{31} = \frac{a_3}{a_1}, \quad R_3 \to R_3 - m_{31} R_1$$

Step 2 — clear the y-column. Using the new second pivot b₂′, subtract a multiple of row 2 from row 3:

$$m_{32} = \frac{b_3'}{b_2'}, \quad R_3 \to R_3 - m_{32} R_2$$

The matrix is now upper-triangular:

$$\left[\begin{array}{ccc|c} a_1 & b_1 & c_1 & d_1 \\ 0 & b_2' & c_2' & d_2' \\ 0 & 0 & c_3'' & d_3'' \end{array}\right]$$

Back substitution

The last row now involves only z, so it is solved directly and the values propagate upward:

$$z = \frac{d_3''}{c_3''}, \quad y = \frac{d_2' - c_2' z}{b_2'}, \quad x = \frac{d_1 - b_1 y - c_1 z}{a_1}$$

Because the matrix is triangular, its determinant is simply the product of the diagonal pivots, $D = a_1 \cdot b_2' \cdot c_3''$. A non-zero product confirms a unique solution.

Worked example

Solve the default system:

$$2x + y - z = 8 \\ -3x - y + 2z = -11 \\ -2x + y + 2z = -3$$

Step 1 — multipliers and first sweep. With pivot a₁ = 2:

$$m_{21} = \frac{-3}{2} = -\tfrac{3}{2}, \qquad m_{31} = \frac{-2}{2} = -1$$

Subtracting these multiples of row 1 gives

$$\left[\begin{array}{ccc|c} 2 & 1 & -1 & 8 \\ 0 & \tfrac{1}{2} & \tfrac{1}{2} & 1 \\ 0 & 2 & 1 & 5 \end{array}\right]$$

Step 2 — clear the y-column. With pivot b₂′ = 1/2:

$$m_{32} = \frac{2}{1/2} = 4, \qquad R_3 \to R_3 - 4 R_2$$ $$\left[\begin{array}{ccc|c} 2 & 1 & -1 & 8 \\ 0 & \tfrac{1}{2} & \tfrac{1}{2} & 1 \\ 0 & 0 & -1 & 1 \end{array}\right]$$

Step 3 — back substitution:

$$z = \frac{1}{-1} = -1, \quad y = \frac{1 - \tfrac{1}{2}(-1)}{1/2} = 3, \quad x = \frac{8 - (1)(3) - (-1)(-1)}{2} = 2$$

The determinant is the product of pivots, $2 \cdot \tfrac{1}{2} \cdot (-1) = -1 \ne 0$, confirming the unique solution (x, y, z) = (2, 3, −1).

Verification:

$$2(2) + 3 - (-1) = 8 \checkmark \quad -3(2) - 3 + 2(-1) = -11 \checkmark \quad -2(2) + 3 + 2(-1) = -3 \checkmark$$

When elimination stalls

Two situations interrupt the clean sweep above:

• Zero pivot, non-singular system. If a coefficient on the diagonal is zero — for example when the first equation has no x-term — the multiplier would divide by zero. The fix is partial pivoting: swap the offending row with one below it that has a non-zero entry in that column, then continue. The solution is unchanged; only the row order differs. Numerical software also swaps in the largest available pivot to limit rounding error.
• Zero pivot that cannot be fixed. If elimination produces a row of zeros across the coefficient columns, the determinant is zero and the coefficient matrix is singular. The system then has either infinitely many solutions (the constants are consistent with the dependency) or none (they contradict it). No single (x, y, z) exists.

Gaussian elimination versus Cramer's rule

Both methods return the same answer for a non-singular system. Cramer's rule expresses each unknown as a ratio of determinants, which is neat for a 3×3 system but becomes costly as the size grows. Gaussian elimination transforms the whole system with row operations and scales far better — it is the backbone of the linear solvers inside numerical libraries. Reducing all the way to reduced row-echelon form (RREF), where each pivot is 1 and the entries above it are also cleared, lets you read the solution directly without a separate back-substitution pass.

Applications

Three-variable systems appear throughout algebra and precalculus, in circuit analysis with Kirchhoff's laws across three loops, in balancing chemical reactions, and wherever three linear conditions must hold simultaneously. Gaussian elimination is the standard hand method for such systems and the conceptual starting point for LU factorization and most computational linear algebra.

Frequently Asked Questions (FAQ)

What is Gaussian elimination?

Gaussian elimination is a systematic way to solve a linear system by combining equations. Writing the system as an augmented matrix, you subtract multiples of one row from the others to create zeros below the diagonal, reaching upper-triangular (row-echelon) form. The last row then involves a single unknown, and back substitution recovers the rest from the bottom up.

How does elimination differ from Cramer's rule?

Both reach the same solution. Cramer's rule writes each unknown as a ratio of determinants, which is compact for a 3×3 system but grows expensive quickly. Gaussian elimination instead transforms the whole system with row operations; it scales far better to larger systems and is the method behind most numerical linear-algebra software. This tool shows the elimination steps, while the determinant is reported as the product of the pivots.

What is the difference between REF and RREF?

Row-echelon form (REF) has zeros below every leading entry, giving an upper-triangular shape you finish by back substitution. Reduced row-echelon form (RREF) goes further: each leading entry is 1 and every entry above it is also cleared, so the solution can be read straight off the matrix. This calculator reduces to REF and then back-substitutes, which is the classic Gaussian-elimination route.

Why does my system show "no unique solution"?

That message appears when the determinant is zero, which means one equation is a linear combination of the others. Elimination then produces a row of zeros in the coefficient columns. Depending on the constants, the system has either infinitely many solutions or none. A unique solution exists only when all three pivots are non-zero.

Why would you swap rows during elimination?

If the pivot you need to divide by is zero, the natural elimination order breaks down even for a system that has a unique solution. Swapping the offending row with one below it (partial pivoting) places a non-zero entry on the diagonal so elimination can continue. Numerical software also swaps rows to pick the largest available pivot, which keeps rounding errors small.